What is the answer to 2^{x+1}=10^{-x} ?

The answer to 2^{x+1}=10^{-x} is x=-(ln(2))/(ln(20))

Solve for x: 2^{x+1}=10^{-x}

The solution is x=-(ln(2))/(ln(20))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2^{x + 1} = 1 0^{- x}

x = - \frac{ln ( 2 )}{ln ( 20 )}

Decimal Notation

x = - 0.23137 \dots

Solution steps

2^{x + 1} = 1 0^{- x}

Apply exponent rules

(x + 1) ln (2) = - x ln (10)

Solve

(x + 1) ln (2) = - x ln (10) : x = - \frac{ln ( 2 )}{ln ( 20 )}

x = - \frac{ln ( 2 )}{ln ( 20 )}

e^{- 9 x} = 3

e^{2 + x} = e^{2 x}

(e^{3})^{2 x} = e^{3} e^{2 x}

e^{7 x} = 6 e^{x}

1 + e^{x} = 0

What is the answer to 2^{x+1}=10^{-x} ?
The answer to 2^{x+1}=10^{-x} is x=-(ln(2))/(ln(20))
Solve for x: 2^{x+1}=10^{-x}
The solution is x=-(ln(2))/(ln(20))